By a famous theorem of Euclid and Euler, every even perfect number can be written as $M_p \cdot 2^{p-1}$ where $M_p=2^p-1$ is a Mersenne prime. Then we have that the last digit of this number is periodic with period $4$, and examining the possibilities we find $0$, $6$, $8$ for the last digit. But if the perfect number is divisible by $5$, then $p \equiv 0 \pmod 4$, which cannot occur; on the other hand, $6$ and $8$ are attainable for $6$ and $28$ respectively, which are easily verified to work.

Let $\sigma(n)$ denote the sum of the natural divisors of n. Note that the first couple perfect numbers are 6 and 28, each of which is even and divisible by exactly 2 primes -- with 2 being one of them. Since each of these can be written in the form $2^k p$, assume that any general even perfect number can be written in the form $n=2^k p$ where $p$ is prime. Then $\sigma(2^k p)=2^{k+1} p$ since $\sigma(n)=2n$. Also note that $\gcd(2^k,p)=1$, and thus $\sigma(2^k p)=\sigma(2^k)\sigma(p)$. Also note that $\sigma(2^k)=1+2+2^2+2^3+\dots+2^k=\displaystyle 2^{k+1}-1$. And $\sigma(p)=p+1$. We obtain that $\displaystyle (2^{k+1}-1)(p+1)=2^{k+1}p$. Then $\displaystyle\frac{p} {p+1}=\frac{2^{k+1}-1} {2^{k+1}}$ and note that on both the left and right-hand sides of the equation, numerator is one less than denominator. Thus a solution for $p$ would be $p=2^{k+1}-1$. Then $n=2^k(2^{k+1}-1)$ where $2^{k+1}-1$ is prime (called a Mersenne prime). We can shift over the $k's$ by one to obtain $n=2^{k-1}(2^k-1)$ where $2^k-1$ is prime; and this only occurs for certain values of which $k$ is prime. Now proceed with the above solution given by Cassius.